# sum-product/product-sum

• Aug 2nd 2012, 12:36 AM
hacker804
sum-product/product-sum
solve the following using sum-product/product-sum identities

$\displaystyle cos20+cos100+cos140=0$

$\displaystyle sin19cos11+sin71sin11=\frac{1}{2}$

$\displaystyle cos20cos40cos60cos180=\frac{1}{16}$

$\displaystyle cos20cos40cos80=\frac{1}{8}$
• Aug 2nd 2012, 09:04 AM
srirahulan
Re: sum-product/product-sum
In this case,

Q1,
$\displaystyle cos20+cos100+cos140=0$
L.H.S
=$\displaystyle 2cos60cos40+cos140$

=$\displaystyle cos40+cos140$

=$\displaystyle 2cos90cos50$

=$\displaystyle 0$ (Since cos90=0)

=R.H.S

Q2,
$\displaystyle sin19cos11+sin71sin11=\frac{1}{2}$
L.H.S
=$\displaystyle \frac{2sin19cos11+2sin71sin11}{2}$

=$\displaystyle \frac{sin30+sin8+cos60-cos82}{2}$

=$\displaystyle \frac{\frac{1}{2}+\frac{1}{2}+sin(90-82)-cos82}{2}$

=$\displaystyle \frac{1+cos82-cos82}{2}$

=$\displaystyle \frac{1}{2}$

=R.H.S

Q3,
I think you mention cos180 or cos 80 I can't get any answer.

Q4,
$\displaystyle cos20cos40cos80=\frac{1}{8}$
L.H.S
=$\displaystyle \frac{2cos20cos40cos80}{2}$

=$\displaystyle \frac{(cos60+cos20)cos80}{2}$

=$\displaystyle \frac{(\frac{1+2cos20}{2})cos80}{2}$

=$\displaystyle \frac{\frac{cos80+cos100+cos60}{2}}{2}$

=$\displaystyle \frac{\frac{cos(180-100)+cos100+cos60}{2}}{2}$

=$\displaystyle \frac{\frac{-cos100+cos100+\frac{1}{2}}{2}}{2}$

=$\displaystyle \frac{1}{8}$

=R.H.S
• Aug 2nd 2012, 09:55 PM
hacker804
Re: sum-product/product-sum
Quote:

$\displaystyle \frac{2cos20cos40cos80}{2}$

Where did the 2 in the numerator and denominator come from???
• Aug 2nd 2012, 11:01 PM
srirahulan
Re: sum-product/product-sum
you want to multiply by 2 to numerator and denominator,This is a way to get your answer quickly.